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In Monty Hall problem generalized to $n$ doors, the Monty hall's problem was generalized to $n$ doors with Monty opening $m$ of them, where $m < n - 1$.

The probability of winning by switching is $\frac{n - 1}{n(n - m - 1)}$. I am confused about what the probability of winning by not switching is in this case. Is it $\frac{1}{n}$, $\frac{1}{n - m}$, or $1- \frac{n - 1}{n(n - m - 1)}$? I think it's $\frac{1}{n}$, but this doesn't seem to incorporate monty opening $M$ doors.

user5965026
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  • $m$ doesn't play a role in the probability for not switching. It does play a role in the probability for switching because it plays a role in counting the unopened, unselected doors. Should stress that the case $m=n-1$ would be radically different. Why? – lulu Mar 05 '21 at 17:48
  • @lulu For that case you get division by zero with the above formula. I wonder why it doesn't give you zero instead, since if monty opens $m = n-1$ doors, then it means the one you picked has the prize. – user5965026 Mar 05 '21 at 17:55
  • If $m<n-1$ then Monty's opening of $m$ doors tells you absolutely nothing about your selection. You already knew that there were at least $n-1$ worthless doors you did not select. But if Monty opens $n$ doors then that tells you a lot of information about your selections, so of course your estimate of its probability changes. – lulu Mar 05 '21 at 18:00

1 Answers1

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The probability of winning by not switching is just the probability of picking the door with the car at your initial guess.

Since there are $n$ doors, the car is behind one unique door and you only choose one door, the probability of winning without switching is just $$\frac{1}{n}.$$

Babado
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  • That makes sense to me, but after Monty opens $m$ doors, couldn't that possibly change our probability of winning by staying put? I.e., shouldn't the probability of winning by staying now be conditioned on knowing Monty opened $m$ doors that didn't have the prize? – user5965026 Mar 05 '21 at 17:28
  • If you initially go for the strategy of not changing your initial choice, then it doesn't matter what Monty does next. – Babado Mar 05 '21 at 17:30
  • But we make a decision whether to switch not after Monty opens the $m$ doors. – user5965026 Mar 05 '21 at 17:31
  • Ok, but the order of decision doesn't really matter. You know that Monty will always open $m$ doors with goats. So whether you decide to switch before or after Monty opens the doors, is the same. – Babado Mar 05 '21 at 17:33
  • I agree the order doesn't matter. What I'm confused about is it seems what this is saying is that $P(\text{winning by not switching} | \text{monty opens m doors}) = P(\text{winning by not switching})$ – user5965026 Mar 05 '21 at 17:43
  • The events ${$winning by not switching$}$ and ${$ Monty opens $m$ doors $}$ are independent. So the probabilities above are indeed the same. Think like this: You pick a door and by doing that (if you don't switch) it is already established whether you win or not. So it doesn't matter what Monty does next, he can $m$ doors, $1$ door, all doors... If you don't switch, this won't affect your reward. – Babado Mar 05 '21 at 17:49
  • In the problem Monty only opens doors he knows have goats and he does not open your door; this does not depend on which door you choose. So $P(\text{winning by not switching} \mid \text{Monty opens }m\text{ doors}) = P(\text{winning by not switching})$ is correct – Henry Mar 05 '21 at 18:12
  • Ah, but the odds are not 1 in 3 if a goat door is removed. The host can not open the car door. – Robert DiGiovanni Mar 13 '22 at 18:07